Question

对于具有通常的四个点a，b，c和d的立方Bézier曲线，

对于给定值t，

如何最优雅地找到切线？

Answer 1

曲线的正切只是它的导数。 Michal使用的参数方程式：

P(t) = (1 - t)^3 * P0 + 3t(1-t)^2 * P1 + 3t^2 (1-t) * P2 + t^3 * P3

应该有

的衍生物

dP(t) / dt =  -3(1-t)^2 * P0 + 3(1-t)^2 * P1 - 6t(1-t) * P1 - 3t^2 * P2 + 6t(1-t) * P2 + 3t^2 * P3

顺便说一句，在你之前的问题中，这似乎是错误的。我相信你在那里使用斜率的二次贝塞尔曲线，而不是立方。

从那里开始，实现一个执行此计算的C函数应该是微不足道的，就像Michal已经为曲线本身提供的那样。

Answer 2

以下是经过全面测试的复制和粘贴代码：

沿曲线绘制近似点，和绘制切线。

bezierInterpolation找到分数

bezierTangent找到切线

以下提供的bezierInterpolation 两个版本：

bezierInterpolation完美无缺

altBezierInterpolation完全相同，但它是以扩展的，非常清晰的，解释性的方式编写的。它使算术更容易理解。

使用这两个例程中的任何一个：结果相同。

在这两种情况下，使用bezierTangent查找切线。（注意：Michal令人难以置信的代码库here。）

还包括如何与drawRect:一起使用的完整示例。

// MBBezierView.m    original BY MICHAL stackoverflow #4058979

#import "MBBezierView.h"



CGFloat bezierInterpolation(
    CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d) {
// see also below for another way to do this, that follows the 'coefficients'
// idea, and is a little clearer
    CGFloat t2 = t * t;
    CGFloat t3 = t2 * t;
    return a + (-a * 3 + t * (3 * a - a * t)) * t
    + (3 * b + t * (-6 * b + b * 3 * t)) * t
    + (c * 3 - c * 3 * t) * t2
    + d * t3;
}

CGFloat altBezierInterpolation(
   CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
    {
// here's an alternative to Michal's bezierInterpolation above.
// the result is absolutely identical.
// of course, you could calculate the four 'coefficients' only once for
// both this and the slope calculation, if desired.
    CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
    CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
    CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
    CGFloat C4 = ( a );

    // it's now easy to calculate the point, using those coefficients:
    return ( C1*t*t*t + C2*t*t + C3*t + C4  );
    }







CGFloat bezierTangent(CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
 {
    // note that abcd are aka x0 x1 x2 x3

/*  the four coefficients ..
    A = x3 - 3 * x2 + 3 * x1 - x0
    B = 3 * x2 - 6 * x1 + 3 * x0
    C = 3 * x1 - 3 * x0
    D = x0

    and then...
    Vx = 3At2 + 2Bt + C         */

    // first calcuate what are usually know as the coeffients,
    // they are trivial based on the four control points:

    CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
    CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
    CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
    CGFloat C4 = ( a );  // (not needed for this calculation)

    // finally it is easy to calculate the slope element,
    // using those coefficients:

    return ( ( 3.0 * C1 * t* t ) + ( 2.0 * C2 * t ) + C3 );

    // note that this routine works for both the x and y side;
    // simply run this routine twice, once for x once for y
    // note that there are sometimes said to be 8 (not 4) coefficients,
    // these are simply the four for x and four for y,
    // calculated as above in each case.
 }







@implementation MBBezierView

- (void)drawRect:(CGRect)rect {
    CGPoint p1, p2, p3, p4;

    p1 = CGPointMake(30, rect.size.height * 0.33);
    p2 = CGPointMake(CGRectGetMidX(rect), CGRectGetMinY(rect));
    p3 = CGPointMake(CGRectGetMidX(rect), CGRectGetMaxY(rect));
    p4 = CGPointMake(-30 + CGRectGetMaxX(rect), rect.size.height * 0.66);

    [[UIColor blackColor] set];
    [[UIBezierPath bezierPathWithRect:rect] fill];
    [[UIColor redColor] setStroke];
    UIBezierPath *bezierPath = [[[UIBezierPath alloc] init] autorelease];   
    [bezierPath moveToPoint:p1];
    [bezierPath addCurveToPoint:p4 controlPoint1:p2 controlPoint2:p3];
    [bezierPath stroke];

    [[UIColor brownColor] setStroke];

 // now mark in points along the bezier!

    for (CGFloat t = 0.0; t <= 1.00001; t += 0.05) {
  [[UIColor brownColor] setStroke];

        CGPoint point = CGPointMake(
            bezierInterpolation(t, p1.x, p2.x, p3.x, p4.x),
            bezierInterpolation(t, p1.y, p2.y, p3.y, p4.y));

            // there, use either bezierInterpolation or altBezierInterpolation,
            // identical results for the position

        // just draw that point to indicate it...
        UIBezierPath *pointPath =
           [UIBezierPath bezierPathWithArcCenter:point
             radius:5 startAngle:0 endAngle:2*M_PI clockwise:YES];
        [pointPath stroke];

        // now find the tangent if someone on stackoverflow knows how
        CGPoint vel = CGPointMake(
            bezierTangent(t, p1.x, p2.x, p3.x, p4.x),
            bezierTangent(t, p1.y, p2.y, p3.y, p4.y));

        // the following code simply draws an indication of the tangent
        CGPoint demo = CGPointMake( point.x + (vel.x*0.3),
                                      point.y + (vel.y*0.33) );
        // (the only reason for the .3 is to make the pointers shorter)
        [[UIColor whiteColor] setStroke];
        UIBezierPath *vp = [UIBezierPath bezierPath];
        [vp moveToPoint:point];
        [vp addLineToPoint:demo];
        [vp stroke];
    }   
}

@end

to draw that class...
MBBezierView *mm = [[MBBezierView alloc]
                     initWithFrame:CGRectMake(400,20, 600,700)];
[mm setNeedsDisplay];
[self addSubview:mm];

以下两个例程用于计算近似等距点，以及那些的切线，沿着贝塞尔曲线。

为了清晰和可靠，这些例程以最简单，最具说明性的方式编写。

CGFloat bezierPoint(CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
    {
    CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
    CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
    CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
    CGFloat C4 = ( a );

    return ( C1*t*t*t + C2*t*t + C3*t + C4  );
    }

CGFloat bezierTangent(CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
    {
    CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
    CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
    CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
    CGFloat C4 = ( a );

    return ( ( 3.0 * C1 * t* t ) + ( 2.0 * C2 * t ) + C3 );
    }

四个预先计算的值C1 C2 C3 C4有时被称为贝塞尔曲线的系数。（回想一下，b c d通常被称为四个控制点。）

当然，t从0到1，例如每0.05。

只需为X调用这些例程，然后为Y单独调用一次。

希望它有所帮助！

重要事实：

（1）绝对事实：不幸的是，Apple提供的NO方法肯定是从UIBezierPath中提取点。

（2）不要忘记它就像使用 UIBezierPath中的动画一样简单。 Google many examples。

（3）很多人问，“不能使用CGPathApply从UIBezierPath中提取点数吗？”不， CGPathApply是完全不相关的：它只是给你制作任何路径的指示清单（所以，“从这里开始”，“直线绘制到这一点”等等。）

Answer 3

我发现使用提供的公式太容易出错了。太容易错过一个微妙的或错位的支架。

相比之下，维基百科提供了更清晰，更清晰的衍生恕我直言：

...在代码中很容易实现：

3f * oneMinusT * oneMinusT * (p1 - p0)
+ 6f * t * oneMinusT * (p2 - p1)
+ 3f * t * t * (p3 - p2)

（假设您使用您选择的语言配置了vector-minus;问题并没有特别标记为ObjC，而iOS现在有几个langs可用）

Answer 4

在我意识到对于参数方程，（dy / dt）/（dx / dt）= dy / dx

之前，我无法解决任何问题。

Answer 5

这是我的Swift实现。

我通过消除所有多余的数学运算来尽力优化速度。即进行最少次数的数学运算调用。并使用最少数量的乘法（比求和要昂贵得多）。

创建贝塞尔曲线有0个乘法。然后进行3次相乘得到贝塞尔点。和2个乘法来获得贝塞尔曲线的切线。

struct CubicBezier {

    private typealias Me = CubicBezier
    typealias Vector = CGVector
    typealias Point = CGPoint
    typealias Num = CGFloat
    typealias Coeficients = (C: Num, S: Num, M: Num, L: Num)

    let xCoeficients: Coeficients
    let yCoeficients: Coeficients

    static func coeficientsOfCurve(from c0: Num, through c1: Num, andThrough c2: Num, to c3: Num) -> Coeficients
    {
        let _3c0 = c0 + c0 + c0
        let _3c1 = c1 + c1 + c1
        let _3c2 = c2 + c2 + c2
        let _6c1 = _3c1 + _3c1

        let C = c3 - _3c2 + _3c1 - c0
        let S = _3c2 - _6c1 + _3c0
        let M = _3c1 - _3c0
        let L = c0

        return (C, S, M, L)
    }

    static func xOrYofCurveWith(coeficients coefs: Coeficients, at t: Num) -> Num
    {
        let (C, S, M, L) = coefs
        return ((C * t + S) * t + M) * t + L
    }

    static func xOrYofTangentToCurveWith(coeficients coefs: Coeficients, at t: Num) -> Num
    {
        let (C, S, M, _) = coefs
        return ((C + C + C) * t + S + S) * t + M
    }

    init(from start: Point, through c1: Point, andThrough c2: Point, to end: Point)
    {
        xCoeficients = Me.coeficientsOfCurve(from: start.x, through: c1.x, andThrough: c2.x, to: end.x)
        yCoeficients = Me.coeficientsOfCurve(from: start.y, through: c1.y, andThrough: c2.y, to: end.y)
    }

    func x(at t: Num) -> Num {
        return Me.xOrYofCurveWith(coeficients: xCoeficients, at: t)
    }

    func y(at t: Num) -> Num {
        return Me.xOrYofCurveWith(coeficients: yCoeficients, at: t)
    }

    func dx(at t: Num) -> Num {
        return Me.xOrYofTangentToCurveWith(coeficients: xCoeficients, at: t)
    }

    func dy(at t: Num) -> Num {
        return Me.xOrYofTangentToCurveWith(coeficients: yCoeficients, at: t)
    }

    func point(at t: Num) -> Point {
        return .init(x: x(at: t), y: y(at: t))
    }

    func tangent(at t: Num) -> Vector {
        return .init(dx: dx(at: t), dy: dy(at: t))
    }
}

使用方式：

let bezier = CubicBezier.init(from: .zero, through: .zero, andThrough: .zero, to: .zero)

let point02 = bezier.point(at: 0.2)
let point07 = bezier.point(at: 0.7)

let tangent01 = bezier.tangent(at: 0.1)
let tangent05 = bezier.tangent(at: 0.5)

找到三次贝塞尔曲线上的点的切线

5 个答案: